Difference between revisions of "2015 AIME II Problems/Problem 7"
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==Solution 4== | ==Solution 4== | ||
− | Using the diagram from solution 2 above, label <math>AF</math> to be <math>h</math>. Through Heron's formula, the area of <math>ABC</math> turns out to be <math>90</math>, so using <math>AE</math> as the height and <math>BC</math> as the base yields <math>AE=\frac{36}{5}</math>. Now, through the use of similarity between <math>\triangle APQ</math> and <math>\triangle ABC</math>, you find <math>\frac{w}{25}=\frac{h}{36/5}</math>. Thus, <math>h=\frac{36w}{125}</math>. To find the height of the rectangle, subtract <math>h</math> from <math>\frac{36}{5}</math> to get <math>\left(\frac{36}{5}-\frac{36w}{125}\right)</math>, and multiply this by the other given side <math>w</math> to get <math>\frac{36w}{5}-\frac{36w^2}{125}</math> for the area of the rectangle. Finally, <math>36+125=\boxed{161}</math>. | + | Using the diagram from solution 2 above, label <math>AF</math> to be <math>h</math>. Through Heron's formula, the area of <math>\triangle ABC</math> turns out to be <math>90</math>, so using <math>AE</math> as the height and <math>BC</math> as the base yields <math>AE=\frac{36}{5}</math>. Now, through the use of similarity between <math>\triangle APQ</math> and <math>\triangle ABC</math>, you find <math>\frac{w}{25}=\frac{h}{36/5}</math>. Thus, <math>h=\frac{36w}{125}</math>. To find the height of the rectangle, subtract <math>h</math> from <math>\frac{36}{5}</math> to get <math>\left(\frac{36}{5}-\frac{36w}{125}\right)</math>, and multiply this by the other given side <math>w</math> to get <math>\frac{36w}{5}-\frac{36w^2}{125}</math> for the area of the rectangle. Finally, <math>36+125=\boxed{161}</math>. |
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=6|num-a=8}} | {{AIME box|year=2015|n=II|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:13, 17 July 2015
Problem
Triangle has side lengths , , and . Rectangle has vertex on , vertex on , and vertices and on . In terms of the side length , the area of can be expressed as the quadratic polynomial
Area() = .
Then the coefficient , where and are relatively prime positive integers. Find .
Solution 1
If , the area of rectangle is , so
and . If , we can reflect over PQ, over , and over to completely cover rectangle , so the area of is half the area of the triangle. Using Heron's formula, since ,
so
and
so the answer is .
Solution 2
Similar triangles can also solve the problem.
First, solve for the area of the triangle. . This can be done by Heron's Formula or placing an right triangle on and solving. (The side would be collinear with line )
After finding the area, solve for the altitude to . Let be the intersection of the altitude from and side . Then . Solving for using the Pythagorean Formula, we get . We then know that .
Now consider the rectangle . Since is collinear with and parallel to , is parallel to meaning is similar to .
Let be the intersection between and . By the similar triangles, we know that . Since . We can solve for and in terms of . We get that and .
Let's work with . We know that is parallel to so is similar to . We can set up the proportion:
. Solving for , .
We can solve for then since we know that and .
Therefore, .
This means that .
Solution 3
Heron's Formula gives so the altitude from to has length
Now, draw a parallel to from , intersecting at . Then in parallelogram , and so . Clearly, and are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so Solving gives , so the answer is .
Solution 4
Using the diagram from solution 2 above, label to be . Through Heron's formula, the area of turns out to be , so using as the height and as the base yields . Now, through the use of similarity between and , you find . Thus, . To find the height of the rectangle, subtract from to get , and multiply this by the other given side to get for the area of the rectangle. Finally, .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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